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vova2212 [387]
3 years ago
10

Similar triangles and polygons worksheet help with number 1

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
Yesyesyesyes yesyeso online
You might be interested in
Can someone tell me how to solve this?
liberstina [14]

These two angles are equivalent because they're corresponding angles created by two parallel lines.

Since they're equivalent, we can set them equal to each other.

23x - 5 = 21x + 5

Now solve for x

23x - 5 = 21x + 5

Add 5 to both sides

23x = 21x + 10

Subtract 21x from both sides

2x = 10

Divide both sides by 2

x = 5

Now, plug in 5 as "x" in one of the equations

Angle 1 = (23*5) - 5

= 115 - 5

= 110 degrees

Since they're equivalent, Angle 2 should also be 110 degrees, but let's check in case.

Angle 2 = (21*5) + 5

= 105 + 5

= 110 degrees

4 0
3 years ago
The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th
Feliz [49]

Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{30.9 - 29}{0.8341}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

X = 27.1

Z = \frac{X - \mu}{s}

Z = \frac{27.1 - 29}{0.8341}

Z = -2.28

Z = -2.28 has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

7 0
4 years ago
Help me someone please
Fynjy0 [20]
The first one. The x values repeat
7 0
3 years ago
Read 2 more answers
Divide. −38÷−67 please help dudes
Eva8 [605]

Answer: -7/16

Step-by-step explanation: You flip the right side so it's -3/8*7/6, cross out 3 and 6, so it turns into -1/8*7/2, so 7*(-1) and -8(2), so -7/16

5 0
3 years ago
23,24 and 25 please! Thank you!!
Brut [27]

Answer:

it is too hard, You can caculator with each part.

example:

23, atrigle

24, cricle

25 atrigle

Step-by-step explanation:

3 0
3 years ago
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